A116952 -id:A116952 - cp's OEIS frontend

A134931 a(n) = (5*3^n-3)/2.

1, 6, 21, 66, 201, 606, 1821, 5466, 16401, 49206, 147621, 442866, 1328601, 3985806, 11957421, 35872266, 107616801, 322850406, 968551221, 2905653666, 8716961001, 26150883006, 78452649021, 235357947066, 706073841201, 2118221523606

Offset: 0

Rolf Pleisch, Jan 29 2008

Numbers n where the recurrence s(0)=1, if s(n-1) >= n then s(n) = s(n-1) - n else s(n) = s(n-1) + n produces s(n)=0. - Hugo Pfoertner, Jan 05 2012
A046901(a(n)) = 1. - Reinhard Zumkeller, Jan 31 2013
Binomial transform of A146523: (1, 5, 10, 20, 40, ...) and double binomial transform of A010685: (1, 4, 1, 4, 1, 4, ...). - Gary W. Adamson, Aug 25 2016
Also the number of maximal cliques in the (n+1)-Hanoi graph. - Eric W. Weisstein, Dec 01 2017
a(n) is the least k such that f(a(n-1)+1) + ... + f(k) > f(a(n-2)+1) + ... + f(a(n-1)) for n > 1, where f(n) = 1/(n+1). Because Sum_{k=1..5*3^(n-1)} 1/(a(n)+3*k-1) + 1/(a(n)+3*k) + 1/(a(n)+3*k+1) - 1/((a(n)+1+5*3^n)*5*3^(n-1)) < Sum_{k=1..5*3^(n-1)} 1/(a(n-1)+k+1) < Sum_{k=1..5*3^(n-1)} 1/(a(n)+3*k-1) + 1/(a(n)+3*k) + 1/(a(n)+3*k+1), we have 1 < 1/3 + 1/4 + ... + 1/7 < 1/8 + 1/9 + ... + 1/22 < ... . - Jinyuan Wang, Jun 15 2020

Vincenzo Librandi, Table of n, a(n) for n = 0..500
Eric Weisstein's World of Mathematics, Hanoi Graph
Eric Weisstein's World of Mathematics, Maximal Clique
Index entries for linear recurrences with constant coefficients, signature (4,-3).

Cf. A003462, A007051, A034472, A024023, A067771, A029858.

Cf. A005030, A010685, A046901, A060816, A116952, A146523, A225918.

[(5*3^n-3)/2: n in [0..30]]; // Vincenzo Librandi, Jun 05 2011

seq((5*3^n-3)/2, n= 0..25); # Gary Detlefs, Jun 22 2010

a=1; lst={a}; Do[a=a*3+3; AppendTo[lst,a], {n,0,100}]; lst (* Vladimir Joseph Stephan Orlovsky, Dec 25 2008 *)
Table[(5 3^n - 9)/6, {n, 20}] (* Eric W. Weisstein, Dec 01 2017 *)
(5 3^Range[20] - 9)/6 (* Eric W. Weisstein, Dec 01 2017 *)
LinearRecurrence[{4, -3}, {1, 6}, 20] (* Eric W. Weisstein, Dec 01 2017 *)
CoefficientList[Series[(1 + 2 x)/(1 - 4 x + 3 x^2), {x, 0, 20}], x] (* Eric W. Weisstein, Dec 01 2017 *)

a(n) = (5*3^n-3)/2; /* Joerg Arndt, Apr 14 2013 */

a(n) = 3*(a(n-1) + 1), with a(0)=1.
From R. J. Mathar, Jan 31 2008: (Start)
O.g.f.: (5/2)/(1-3*x) - (3/2)/(1-x).
a(n) = (A005030(n) - 3)/2. (End)
a(n) = A060816(n+1) - 1. - Philippe Deléham, Apr 14 2013
E.g.f.: exp(x)*(5*exp(2*x) - 3)/2. - Stefano Spezia, Aug 28 2023

More terms from Vladimir Joseph Stephan Orlovsky, Dec 25 2008

A237930 a(n) = 3^(n+1) + (3^n-1)/2.

Original entry on oeis.org

3, 10, 31, 94, 283, 850, 2551, 7654, 22963, 68890, 206671, 620014, 1860043, 5580130, 16740391, 50221174, 150663523, 451990570, 1355971711, 4067915134, 12203745403, 36611236210, 109833708631, 329501125894, 988503377683, 2965510133050, 8896530399151

Offset: 0

Philippe Deléham, Feb 16 2014

a(n-1) agrees with the graph radius of the n-Sierpinski carpet graph for n = 2 to at least n = 5. See A100774 for the graph diameter of the n-Sierpinski carpet graph.
The inverse binomial transform gives 3, 7, 14, 28, 56, ... i.e., A005009 with a leading 3. - R. J. Mathar, Jan 08 2020
First differences of A108765. The digital root of a(n) for n > 1 is always 4. a(n) is never divisible by 7 or by 12. a(n) == 10 (mod 84) for odd n. a(n) == 31 (mod 84) for even n > 0. Conjecture: This sequence contains no prime factors p == {11, 13, 23, 61 71, 73} (mod 84). - Klaus Purath, Apr 13 2020
This is a subsequence of A017209 for n > 1. See formula. - Klaus Purath, Jul 03 2020

			Ternary....................Decimal
10...............................3
101.............................10
1011............................31
10111...........................94
101111.........................283
1011111........................850
10111111......................2551
101111111.....................7654, etc.

Colin Barker, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Graph Radius.
Eric Weisstein's World of Mathematics, Sierpinski Carpet Graph.
Index entries for linear recurrences with constant coefficients, signature (4,-3).

Cf. A000244, A003462, A005009, A005032 (first differences), A017209, A060816, A100774, A108765 (partial sums), A199109, A329774.

Cf. A024023, A029858, A057198, A116952, A171498.

[3^(n+1) + (3^n-1)/2: n in [0..40]]; // Vincenzo Librandi, Jan 09 2020

(* Start from Eric W. Weisstein, Mar 13 2018 *)
Table[(7 3^n - 1)/2, {n, 0, 20}]
(7 3^Range[0, 20] - 1)/2
LinearRecurrence[{4, -3}, {10, 31}, {0, 20}]
CoefficientList[Series[(3 - 2 x)/((x - 1) (3 x - 1)), {x, 0, 20}], x]
(* End *)

Vec((3 - 2*x) / ((1 - x)*(1 - 3*x)) + O(x^30)) \\ Colin Barker, Nov 27 2019

G.f.: (3-2*x)/((1-x)*(1-3*x)).
a(n) = A000244(n+1) + A003462(n).
a(n) = 3*a(n-1) + 1 for n > 0, a(0)=3. (Note that if a(0) were 1 in this recurrence we would get A003462, if it were 2 we would get A060816. - N. J. A. Sloane, Dec 06 2019)
a(n) = 4*a(n-1) - 3*a(n-2) for n > 1, a(0)=3, a(1)=10.
a(n) = 2*a(n-1) + 3*a(n-2) + 2 for n > 1.
a(n) = A199109(n) - 1.
a(n) = (7*3^n - 1)/2. - Eric W. Weisstein, Mar 13 2018
From Klaus Purath, Apr 13 2020: (Start)
a(n) = A057198(n+1) + A024023(n).
a(n) = A029858(n+2) - A024023(n).
a(n) = A052919(n+1) + A029858(n+1).
a(n) = (A000244(n+1) + A171498(n))/2.
a(n) = 7*A003462(n) + 3.
a(n) = A116952(n) + 2. (End)
a(n) = A017209(7*(3^(n-2)-1)/2 + 3), n > 1. - Klaus Purath, Jul 03 2020
E.g.f.: exp(x)*(7*exp(2*x) - 1)/2. - Stefano Spezia, Aug 28 2023

A225918 a(n) is the least k such that f(a(n-1)+1) + ... + f(k) > f(a(n-2)+1) + ... + f(a(n-1)) for n > 1, where f(n) = 1/(n+3) and a(1) = 1.

Original entry on oeis.org

1, 9, 32, 98, 287, 828, 2377, 6812, 19510, 55866, 159958, 457987, 1311283, 3754381, 10749290, 30776629, 88117519, 252291984, 722344942, 2068168017, 5921435438, 16953843853, 48541071558, 138979434294, 397916291012, 1139286366040, 3261925819973, 9339320097349, 26739694491713

Offset: 1

Clark Kimberling, May 21 2013

nonn

Suppose that f(n) is a sequence of positive real numbers for which the series f(1) + f(2) + ... diverges. Put a(1) = 1 and a(n) = least k such that f(a(n-1)+1) + ... + f(k) > f(a(n-2)+1) + ... + f(a(n-1)) for n > 1. Conjecture: a(n) is linearly recurrent for the choices of f(n) shown here:
f(n) ...... a(n)................ recurrence coefficients
1/n ....... A003462: 1,4,13,.... (4,-3)
1/(n+1) ... A134931: 1,6,21,.... (4,-3)
1/(n+2) ... A116952: 1,8,29,.... (4,-3)
1/(n+3) ... A225918: 1,9,32,.... (3,0,-1,0,-1)
1/(n+4) ... A225919: 1,11,40,... (4,-4,3,-2)
1/(n+5) ... A225920: 1,13,48,... ?
1/(n+6) ... A225921: 1,14,50,... ?
1/(n+7) ... A225922: 1,16,48,... ?
Assuming linear recurrence, it appears that lim_{n->infinity} a(n+1)/a(n) is the greatest root, R, of the characteristic polynomial of the recurrence, and that lim_{n->infinity} (1/(a(n-1)+1) + ... + 1/a(n)) = log R.
For sequences A225920-A225922, linear recurrence is unlikely to exist. - Max Alekseyev, Jan 27 2022

			a(1) = 1 by decree; a(2) = 9 because 1/5 + ... + 1/11 < 1 < 1/5 + ... + 1/(9+3), so that a(3) = 32 because 1/13 + ... + 1/34 < 1/5 + ... + 1/12 < 1/13 + ... + 1/(32+3).
Successive values of a(n) yield a chain: 1 < 1/(1+4) + ... + 1/(9+3) < 1/(9+4) + ... + 1/(32+3) < 1/(32+4) + ... + 1/(98+3) < ...
Abbreviating this chain as b(1) = 1 < b(2) < b(3) < b(4) < ... < R = 2.8631..., it appears that lim_{n->infinity} b(n) = log R = 1.0519... .

Cf. A003462, A134931, A116952.

Cf. A225919, A225920, A225921, A225922.

nn = 11; f[n_] := 1/(n + 3); a[1] = 1; g[n_] := g[n] = Sum[f[k], {k, 1, n}]; s = 0; a[2] = NestWhile[# + 1 &, 2, ! (s += f[#]) >= a[1] &]; s = 0; a[3] = NestWhile[# + 1 &, a[2] + 1, ! (s += f[#]) >= g[a[2]] - f[1] &]; Do[s = 0; a[z] = NestWhile[# + 1 &, a[z - 1] + 1, ! (s += f[#]) >= g[a[z - 1]] - g[a[z - 2]] &], {z, 4, nn}]; m = Map[a, Range[nn]] (* Peter J. C. Moses, May 13 2013 *)

lista(nn) = {default(realprecision, 100); my(k=5, r=1, s); print1(1); for(n=2, nn, s=0; while((s+=1./k)Jinyuan Wang, Jun 14 2020

For n>=3, a(n) = ceiling( (a(n-1)+3.5)^2 / (a(n-2)+3.5) - 3.5 ) unless the fractional part of the number inside ceiling() is very small (~ 1/a(n-2)). - Max Alekseyev, Jan 27 2022

a(12)-a(18) from Robert G. Wilson v, May 22 2013
a(19) from Jinyuan Wang, Jun 14 2020
Terms a(20) on from Max Alekseyev, Jan 27 2022

A117617 a(n) = 5*a(n-1) + 3 with a(0) = 1.

Original entry on oeis.org

1, 8, 43, 218, 1093, 5468, 27343, 136718, 683593, 3417968, 17089843, 85449218, 427246093, 2136230468, 10681152343, 53405761718, 267028808593, 1335144042968, 6675720214843, 33378601074218, 166893005371093

Offset: 0

Parthasarathy Nambi, Apr 07 2006

nonn

			If n=1 then a(1) = 5*a(0) + 3 = 5*1 + 3 = 8 which is the second term.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6, -5).

Cf. A116952.

a[0] := 1; a[n_] := 5*a[n - 1] + 3; Table[a[n], {n, 0, 35}] (* Stefan Steinerberger, Apr 08 2006 *)
NestList[5#+3&,1,20] (* Harvey P. Dale, Jun 14 2024 *)

G.f.: (1+2*x)/((1-x)*(1-5*x)). - Philippe Deléham, Feb 22 2014

a(n) = 6*a(n-1) - 5*a(n-2), a(0) = 1, a(1) = 8. - Philippe Deléham, Feb 22 2014

More terms from Stefan Steinerberger, Apr 08 2006

A117088 a(n) = (11*5^n - 7)/4.

Original entry on oeis.org

1, 12, 67, 342, 1717, 8592, 42967, 214842, 1074217, 5371092, 26855467, 134277342, 671386717, 3356933592, 16784667967, 83923339842, 419616699217, 2098083496092, 10490417480467, 52452087402342, 262260437011717, 1311302185058592, 6556510925292967

Offset: 0

Parthasarathy Nambi, Apr 17 2006

			If n=1 then 5*(n-1) + 7 = 5*1 + 7 = 12, which is the second term.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-5).

Cf. A117617, A116952.

[(11*5^n-7)/4 : n in [0..25]]; // Wesley Ivan Hurt, Nov 12 2014

A117088:=n->(11*5^n-7)/4: seq(A117088(n), n=0..25); # Wesley Ivan Hurt, Nov 12 2014

NestList[5# + 7 &, 1, 50] (* Stefan Steinerberger, Apr 21 2006 *)

a(n) = 5*a(n-1) + 7 with a(0) = 1.
G.f.: (1+6*x)/((1-x)*(1-5*x)). - Philippe Deléham, Feb 22 2014
a(n) = 6*a(n-1) - 5*a(n-2), a(0) = 1, a(1) = 12. - Philippe Deléham, Feb 22 2014
a(n) = (11*5^n - 7) / 4. - Ralf Stephan, Feb 23 2014

Definition corrected and better name by Ralf Stephan, Feb 23 2014

A137215 a(n) = 3(10^n) + (n^2 + 1)(10^n - 1)/9.

Original entry on oeis.org

3, 32, 355, 4110, 48887, 588886, 7111107, 85555550, 1022222215, 12111111102, 142222222211, 1655555555542, 19111111111095, 218888888888870, 2488888888888867, 28111111111111086, 315555555555555527, 3522222222222222190, 39111111111111111075, 432222222222222222182

Offset: 0

Ctibor O. Zizka, Mar 06 2008

Sequence generalized: a(n) = a(0)*(B^n) + F(n)* ((B^n)-1)/(B-1); a(0), B integers, F(n) arithmetic function.
Examples:
a(0) = 1, B = 10, F(n) = 1 gives A002275, F(n) = 2 gives A090843, F(n) = 3 gives A097166, F(n) = 4 gives A099914, F(n) = 5 gives A099915.
a(0) = 1, B = 2, F(n) = 1 gives A000225, F(n) = 2 gives A033484, F(n) = 3 gives A036563, F(n) = 4 gives A048487, F(n) = 5 gives A048488, F(n) = 6 gives A048489.
a(0) = 1, B = 3, F(n) = 1 gives A003462, F(n) = 2 gives A048473, F(n) = 3 gives A134931, F(n) = 4 gives A058481, F(n) = 5 gives A116952.
a(0) = 1, B = 4, F(n) = 1 gives A002450, F(n) = 2 gives A020989, F(n) = 3 gives A083420, F(n) = 4 gives A083597, F(n) = 5 gives A083584.
a(0) = 1, B = 5, F(n) = 1 gives A003463, F(n) = 2 gives A057651, F(n) = 3 gives A117617, F(n) = 4 gives A081655.
a(0) = 2, B = 10, F(n) = 1 gives A037559, F(n) = 2 gives A002276.

			a(3) = 3*10^3 + (3*3 + 1)*(10^3 - 1)/9 = 4110.

G. C. Greubel, Table of n, a(n) for n = 0..985
Index entries for linear recurrences with constant coefficients, signature (33,-393,1991,-3930,3300,-1000).

Cf. A002275, A011557.

Table[3*10^n +(n^2 +1)*(10^n -1)/9, {n,0,30}] (* G. C. Greubel, Jan 05 2022 *)

a(n) = 3*(10^n) + (n*n+1)*((10^n)-1)/9; \\ Jinyuan Wang, Feb 27 2020

[3*10^n +(1+n^2)*(10^n -1)/9 for n in (0..30)] # G. C. Greubel, Jan 05 2022

a(n) = 3*(10^n) + (n^2 + 1)*(10^n - 1)/9.

O.g.f.: (3 - 67*x + 478*x^2 - 1002*x^3 + 850*x^4 - 100*x^5)/((1-x)^3 * (1-10*x)^3). - R. J. Mathar, Mar 16 2008

More terms from R. J. Mathar, Mar 16 2008

More terms from Jinyuan Wang, Feb 27 2020

cp's OEIS Frontend

A134931 a(n) = (5*3^n-3)/2.

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A237930 a(n) = 3^(n+1) + (3^n-1)/2.

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A225918 a(n) is the least k such that f(a(n-1)+1) + ... + f(k) > f(a(n-2)+1) + ... + f(a(n-1)) for n > 1, where f(n) = 1/(n+3) and a(1) = 1.

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A117617 a(n) = 5*a(n-1) + 3 with a(0) = 1.

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A117088 a(n) = (11*5^n - 7)/4.

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A137215 a(n) = 3*(10^n) + (n^2 + 1)*(10^n - 1)/9.

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A137215 a(n) = 3(10^n) + (n^2 + 1)(10^n - 1)/9.